What is the derivative of 1/4 x^4+\sqrt[3]{x^3} ?

The derivative of 1/4 x^4+\sqrt[3]{x^3} is x^3+1

What is the first derivative of 1/4 x^4+\sqrt[3]{x^3} ?

The first derivative of 1/4 x^4+\sqrt[3]{x^3} is x^3+1

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x^{3} + 1

Solution steps

Simplify

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

\frac{d}{d x} (\frac{x ^{4}}{4}) = x^{3}

= x^{3} + 1

arcsec (x) \cdot arccos (x)

\int - y^{2} d y

\frac{d}{d x} (\frac{- 1}{4 x ^{\frac{3}{2}}})

n = 0 \sum \infty \frac{1}{2 n}

\int x^{2} - x^{4} d x

What is the derivative of 1/4 x^4+\sqrt[3]{x^3} ?
The derivative of 1/4 x^4+\sqrt[3]{x^3} is x^3+1
What is the first derivative of 1/4 x^4+\sqrt[3]{x^3} ?
The first derivative of 1/4 x^4+\sqrt[3]{x^3} is x^3+1